A balloon with a diameter of 20 m is charged to a potential of 300 kV due to friction against air. With what force does the Shas

A balloon with a diameter of 20 m is charged to a potential of 300 kV due to friction against air. With what force does the Shas interact with the planet (Earth)?

Let’s write down the power of the Coulomb. It will be equal to the force of interaction of the charges of the Earth and the ball, therefore:
Fк = (k | q1 | * | q2 |) / r ^ 2
Fк = Eq
Fк = Fq

Let us write out the potential and express q:
φ = kq / r
φ = 2kq / d
q = φd / 2k

Since Fк = Eq, then we substitute the value of q and calculate the force of interaction between the Earth and the ball:
Fк = Eφd / 2k
Fк = (100 * 300 * 10 ^ 3 * 20) / 2 * 9 * 10 ^ 9 = 1/30 = 0.0333 … (H)
Fк = 33 (mN)

Answer: Fк = 33 (mN)